Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $q = \dfrac{k - 10}{k^2 - 2k - 48} \div \dfrac{-7k + 70}{2k - 16} $
Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{k - 10}{k^2 - 2k - 48} \times \dfrac{2k - 16}{-7k + 70} $ First factor the quadratic. $q = \dfrac{k - 10}{(k - 8)(k + 6)} \times \dfrac{2k - 16}{-7k + 70} $ Then factor out any other terms. $q = \dfrac{k - 10}{(k - 8)(k + 6)} \times \dfrac{2(k - 8)}{-7(k - 10)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ (k - 10) \times 2(k - 8) } { (k - 8)(k + 6) \times -7(k - 10) } $ $q = \dfrac{ 2(k - 10)(k - 8)}{ -7(k - 8)(k + 6)(k - 10)} $ Notice that $(k - 10)$ and $(k - 8)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ 2(k - 10)\cancel{(k - 8)}}{ -7\cancel{(k - 8)}(k + 6)(k - 10)} $ We are dividing by $k - 8$ , so $k - 8 \neq 0$ Therefore, $k \neq 8$ $q = \dfrac{ 2\cancel{(k - 10)}\cancel{(k - 8)}}{ -7\cancel{(k - 8)}(k + 6)\cancel{(k - 10)}} $ We are dividing by $k - 10$ , so $k - 10 \neq 0$ Therefore, $k \neq 10$ $q = \dfrac{2}{-7(k + 6)} $ $q = \dfrac{-2}{7(k + 6)} ; \space k \neq 8 ; \space k \neq 10 $